Question: $ D = \left[\begin{array}{rr}3 & 5 \\ 5 & -1 \\ 3 & -1\end{array}\right]$ $ C = \left[\begin{array}{rr}0 & 1 \\ -1 & 4\end{array}\right]$ What is $ D C$ ?
Answer: Because $ D$ has dimensions $(3\times2)$ and $ C$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(3\times2)$ $ D C = \left[\begin{array}{rr}{3} & {5} \\ {5} & {-1} \\ \color{gray}{3} & \color{gray}{-1}\end{array}\right] \left[\begin{array}{rr}{0} & \color{#DF0030}{1} \\ {-1} & \color{#DF0030}{4}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ D$ , with the corresponding elements in column $j$ of the second matrix, $ C$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ D$ with the first element in ${\text{column }1}$ of $ C$ , then multiply the second element in ${\text{row }1}$ of $ D$ with the second element in ${\text{column }1}$ of $ C$ , and so on. Add the products together. $ \left[\begin{array}{rr}{3}\cdot{0}+{5}\cdot{-1} & ? \\ ? & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ D$ with the corresponding elements in ${\text{column }1}$ of $ C$ and add the products together. $ \left[\begin{array}{rr}{3}\cdot{0}+{5}\cdot{-1} & ? \\ {5}\cdot{0}+{-1}\cdot{-1} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ D$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ C$ and add the products together. $ \left[\begin{array}{rr}{3}\cdot{0}+{5}\cdot{-1} & {3}\cdot\color{#DF0030}{1}+{5}\cdot\color{#DF0030}{4} \\ {5}\cdot{0}+{-1}\cdot{-1} & ? \\ ? & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{3}\cdot{0}+{5}\cdot{-1} & {3}\cdot\color{#DF0030}{1}+{5}\cdot\color{#DF0030}{4} \\ {5}\cdot{0}+{-1}\cdot{-1} & {5}\cdot\color{#DF0030}{1}+{-1}\cdot\color{#DF0030}{4} \\ \color{gray}{3}\cdot{0}+\color{gray}{-1}\cdot{-1} & \color{gray}{3}\cdot\color{#DF0030}{1}+\color{gray}{-1}\cdot\color{#DF0030}{4}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}-5 & 23 \\ 1 & 1 \\ 1 & -1\end{array}\right] $